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\title{\textbf{2025微分几何期末A卷}}
\author{唐嘉琪}
\date{\today}
\linespread{1.25  }
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\newenvironment{problem}{\begin{shaded}\stepcounter{problemname}\par\noindent\textbf{题目\arabic{problemname}. }}{\end{shaded}\par}
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\newenvironment{note}{\par\noindent\textbf{题目\arabic{problemname}的注记. }}{\par}
\newtheorem{theorem}{定理}
\newtheorem{definition}{定义}
\newtheorem{lemma}{引理}


\begin{document}

\maketitle

\begin{problem}
	试证明:
	\begin{itemize}
		\item 实$1$维单位球面$\mathbb S^1$为一个实$1$维光滑流形.
		\item 包含映射$i\colon \mathbb S^1\hookrightarrow\mathbb R^2$为嵌入, 其中$\mathbb R^2$为实$2$维欧氏空间.
	\end{itemize}
\end{problem}

\begin{solution}
这里给出更一般的证明, 试题只是$m=1$的特例.
\begin{itemize}
	\item $m$维单位球面$\mathbb S^m$定义为
	\begin{align*}
		\mathbb S^m:=\left\{(a_1,\cdots,a_{m+1})\in\mathbb R\colon \sum_{i=1}^{m+1}a_i^2=1\right\}.
	\end{align*}
	
	令$p=(0,\cdots,0,1)$为北极, $q=(0,\cdots,0,-1)$为南极. 开集$U(p)=\mathbb S^m\setminus \{p\}$和$U(q)=\mathbb S^m\setminus \{q\}$之并覆盖$\mathbb S^m$. 下面定义映射$\varphi$ 和$\psi$, 使得$(U(p),\varphi)$和$(U(q),\psi)$是覆盖$\mathbb S^m$的两个$C^\infty$坐标图. 映射$\varphi$和$\psi$由球极投影确定.
	
	对$\forall a\in U(p)$, 记$\lambda$是由点$p$和$a$确定的直线, $\pi$是$\mathbb R^{m+1}$内由$x_{n+1}=0$确定的超平面, $\pi(a)$表示$\mathbb R^{m+1}$内直线$\lambda$和$\pi$相交的点. 当$a=(a_1,\cdots,a_{m+1})$时, 易得$\pi(a)=(x_1,\cdots,x_m,0)$, 这里$x_i=a_i/(1-a_{m+1})$ ($1\leqslant i\leqslant m$). 令
	\begin{align*}
		\varphi(a)=(x_1,\cdots,x_m)=\left(\dfrac{a_1}{1-a_{m+1}},\cdots,\dfrac{a_m}{1-a_{m+1}}\right),
	\end{align*}
	$\varphi$是$U(p)$到$\mathbb R^m$上的一个同胚. 因为如果$(x_1,\cdots,x_m)$已知, 则由计算可得
	\begin{align*}
		a=\left(\dfrac{2x_1}{1+\sum\limits_{i=1}^mx_i^2},\dfrac{2x_2}{1+\sum\limits_{i=1}^mx_i^2},\cdots,\dfrac{2x_m}{1+\sum\limits_{i=1}^mx_i^2},\dfrac{\sum\limits_{i=1}^mx_i^2-1}{1+\sum\limits_{i=1}^mx_i^2}\right).
	\end{align*}
	
	类似的, 定义
	\begin{align*}
		\psi\colon U(q)\to \mathbb R^m,\quad (a_1,\cdots,a_{m+1})\mapsto(y_1,\cdots,y_m),
	\end{align*}
	这里$y_i=a_i/(1+a_{m+1})$. 如果$(y_1,\cdots,y_m)$已知, 则
	\begin{align*}
		a=\left(\dfrac{2y_1}{1+\sum\limits_{i=1}^my_i^2},\dfrac{2y_2}{1+\sum\limits_{i=1}^my_i^2},\cdots,\dfrac{2y_m}{1+\sum\limits_{i=1}^my_i^2},\dfrac{1-\sum\limits_{i=1}^my_i^2}{1+\sum\limits_{i=1}^my_i^2}\right).
	\end{align*}
	由于
	\begin{align*}
		U(p)\cap U(q)=\mathbb S^m\setminus(\{p\}\cup\{q\}),\\
		\forall a\in U(p)\cap U(q),\quad \varphi(a)=(x_1,\cdots,x_m),\\
		\psi(a)=(y_1,\cdots,y_m),
	\end{align*}
	而
	\begin{align*}
		y_i=\dfrac{a_i}{1+a_{m+1}}=\dfrac{x_i}{\sum\limits_{i=1}^mx_i^2},\quad x_i=\dfrac{a_i}{1-a_{m+1}}=\dfrac{y_i}{\sum\limits_{i=1}^my_i^2},
	\end{align*}
	因此$\psi\varphi^{-1}$和$\varphi\psi^{-1}$都是$C^\infty$的, 于是$\mathbb S^m$是一个$m$维光滑流形.

	\item 只要证明$i$是一个一一浸入, 且是一个到内的同胚即可.
	
	我们继续沿用上一题的记号, 在球极投影下有(我们选择一个, 另一个是一样的)
	\begin{align*}
		i(x^1,\cdots,x^m)=\left(\dfrac{2x_1}{1+\sum\limits_{i=1}^mx_i^2},\dfrac{2x_2}{1+\sum\limits_{i=1}^mx_i^2},\cdots,\dfrac{2x_m}{1+\sum\limits_{i=1}^mx_i^2},\dfrac{\sum\limits_{i=1}^mx_i^2-1}{1+\sum\limits_{i=1}^mx_i^2}\right).
	\end{align*}
	在不会引起歧义的时候, 为了方便, 后文将$\sum\limits_{i=1}^m$简写为$\sum$.
	
	考察切映射$i_{\star,s}\colon \operatorname T_{\mathbb S^m,s}\to \operatorname T_{\mathbb R^{m+1},i(s)}$的Jacobi阵.
	
	先计算前面$m$个分量的偏导数,
	\begin{align*}
		\dfrac{\partial}{\partial x_k}\left(\dfrac{2x_j}{1+\sum x_i^2}\right)_{\substack{1\leqslant j\leqslant m\\ 1\leqslant k\leqslant m}}.
	\end{align*}
	当$j=k$时,
	\begin{align*}
		\dfrac{\partial}{\partial x_j}\left(\dfrac{2x_j}{1+\sum x_i^2}\right)=\dfrac{2(1+\sum x_i^2)-2x_j\cdot 2x_j}{(1+\sum x_i^2)^2}=\dfrac{2(1+\sum x_i^2-2x_j^2)}{(1+\sum x_i^2)^2}.
	\end{align*}
	当$j\neq k$时,
	\begin{align*}
		\dfrac{\partial}{\partial x_k}\left(\dfrac{2x_j}{1+\sum x_i^2}\right)=2x_j\dfrac{-2x_k}{(1+\sum x_i^2)^2}=\dfrac{-4x_jx_k}{(1+\sum x_i^2)^2}.
	\end{align*}
	
	再计算第$(m+1)$个分量的偏导数,
	\begin{align*}
		\dfrac{\partial}{\partial x_k}\left(\dfrac{\sum x_i^2-1}{1+\sum x_i^2}\right)_{1\leqslant k\leqslant m}=\dfrac{2x_k(1+\sum x_i^2)+(1-\sum x_i^2)\cdot 2x_k}{(1+\sum x_i^2)^2}=\dfrac{4x_k}{(1+\sum x_i^2)^2}
	\end{align*}
	
	现在我们来看$\operatorname{Jac}(i_{\star,s})$
	\begin{align*}
		\operatorname{Jac}(i_{\star,s})=\begin{bmatrix}
			\dfrac{\partial}{\partial x_1}\left(\dfrac{2x_1}{1+\sum x_i^2}\right) & \dfrac{\partial}{\partial x_2}\left(\dfrac{2x_1}{1+\sum x_i^2}\right) & \cdots & \dfrac{\partial}{\partial x_1}\left(\dfrac{2x_m}{1+\sum x_i^2}\right)\\[0.4cm]
			\dfrac{\partial}{\partial x_1}\left(\dfrac{2x_2}{1+\sum x_i^2}\right) & \dfrac{\partial}{\partial x_2}\left(\dfrac{2x_2}{1+\sum x_i^2}\right) & \cdots & \dfrac{\partial}{\partial x_m}\left(\dfrac{2x_2}{1+\sum x_i^2}\right)\\
			\vdots & \vdots & \ddots & \vdots \\
			\dfrac{\partial}{\partial x_1}\left(\dfrac{2x_m}{1+\sum x_i^2}\right) & \dfrac{\partial}{\partial x_2}\left(\dfrac{2x_m}{1+\sum x_i^2}\right) & \cdots & \dfrac{\partial}{\partial x_m}\left(\dfrac{2x_m}{1+\sum x_i^2}\right)\\[0.4cm]
			\dfrac{\partial}{\partial x_1}\left(\dfrac{\sum x_i^2-1}{1+\sum x_i^2}\right) & \dfrac{\partial}{\partial x_2}\left(\dfrac{\sum x_i^2-1}{1+\sum x_i^2}\right) & \cdots & \dfrac{\partial}{\partial x_m}\left(\dfrac{\sum x_i^2-1}{1+\sum x_i^2}\right)
		\end{bmatrix}_{(m+1)\times m}\\
		=\dfrac{1}{(1+\sum x_i^2)^2}\begin{bmatrix}
			2(1+\sum x_i^2-2x_1^2) & -4x_1x_2 & \cdots & -4x_1x_m\\
			-4x_2x_1 & 2(1+\sum x_i^2-2x_2^2) & \cdots & -4x_2x_m\\
			\vdots & \vdots & \ddots & \vdots\\
			-4x_mx_1 & -4x_mx_2 & \cdots & 2(1+\sum x_i^2-2x_m^2)\\
			4x_1 & 4x_2 & \cdots & 4x_m
		\end{bmatrix}
	\end{align*}
	我们记上式这个矩阵为$\Lambda$, 即$\operatorname{Jac}(i_{\star,s})=(1+\sum x_i^2)^{-2}\cdot \Lambda$. 对$\Lambda$进行初等变换, 有
	\begin{align*}
		\Lambda\to \begin{bmatrix}
			1 & 0 & \cdots & 0\\
			0 & 1 & \cdots & 0\\
			\vdots & \vdots & \ddots & \vdots \\
			0 & 0 & \cdots & 1\\
			x_1 & x_2 & \cdots & x_m
		\end{bmatrix}
	\end{align*}
	这足以说明$\operatorname{rk}(i_{\star,s})=\operatorname{rk}(\Lambda)=m$, 即$f_{\star,s}$非退化.
	
	$i$的单性是显然的, 这说明了$i$是一个一一浸入, 同时注意到上述球极投影是一个经典的到内同胚, 从而$i\colon \mathbb S^m\hookrightarrow \mathbb R^{m+1}$是一个嵌入.
	\end{itemize}
\end{solution}

\begin{problem}
	假定$M$为实$2$维光滑流形, $f\in C^\infty(M)$为光滑流形$M$上光滑函数, 记$\mathrm d$为外微分算子.试证明: 
	\begin{itemize}
		\item $\mathrm df$与$M$上局部光滑坐标选取无关.
		\item $\mathrm d^2f=0$.
	\end{itemize}
\end{problem}

\begin{solution}
\begin{itemize}
	\item 选取两组光滑坐标$\{x^1,x^2\},\{y^1,y^2\}$.
	\begin{align*}
		\mathrm df=&\dfrac{\partial f}{\partial y^1}\mathrm dy^1+\dfrac{\partial f}{\partial y^2}\mathrm dy^2=\dfrac{\partial f}{\partial y^1}\left(\dfrac{\partial y^1}{\partial x^1}\mathrm dx^1+\dfrac{\partial y^1}{\partial x^2}\mathrm dx^2\right)+\dfrac{\partial f}{\partial y^2}\left(\dfrac{\partial y^2}{\partial x^1}\mathrm dx^1+\dfrac{\partial y^2}{\partial x^2}\mathrm dx^2\right)\\
		=&\left(\dfrac{\partial f}{\partial y^1}\dfrac{\partial y^1}{\partial x^1}+\dfrac{\partial f}{\partial y^2}\dfrac{\partial y^2}{\partial x^1}\right)\mathrm dx^1+\left(\dfrac{\partial f}{\partial y^1}\dfrac{\partial y^1}{\partial x^2}+\dfrac{\partial f}{\partial y^2}\dfrac{\partial y^2}{\partial x^2}\right)\mathrm dx^2=\dfrac{\partial f}{\partial x^1}\mathrm dx^1+\dfrac{\partial f}{\partial x^2}\mathrm dx^2
	\end{align*}
	\item 局部光滑坐标记为$\{x^1,x^2\}$.
	\begin{align*}
		\mathrm d^2f=&\mathrm d\left(\sum_{i=1}^2\dfrac{\partial f}{\partial x^i}\mathrm dx^i\right)=\sum_{i,j=1}^2\dfrac{\partial^2f}{\partial x^j\partial x^i}\mathrm dx^j\wedge\mathrm dx^i=\dfrac{1}{2}\sum_{i,j=1}^2\left(\dfrac{\partial^2 f}{\partial x^j\partial x^i}-\dfrac{\partial^2 f}{\partial x^i\partial x^j}\right)=0
	\end{align*}
\end{itemize}
\end{solution}

\begin{problem}
	假定$D$为$2$维实空间$\mathbb R^2\colon\{x,y\}$上有界区域. $\partial D$为$D$的边界, 其上定向由$D$上定向诱导. 又$P=P(x,y), Q=Q(x,y)$为$D$上光滑函数. 利用流形上Stokes公式, 试证明:
	\begin{align*}
		\int_{\partial D}P\mathrm dx+Q\mathrm dy=\int_D\left(-\dfrac{\partial P}{\partial y}+\dfrac{\partial Q}{\partial x}\right)\mathrm dx\wedge \mathrm dy.
	\end{align*}
\end{problem}
\begin{solution}
	令$\omega=P\mathrm dx+Q\mathrm dy$, 则
	\begin{align*}
		\mathrm d\omega=&\mathrm d(P\mathrm dx+Q\mathrm dy)=\mathrm dP\wedge\mathrm dx+P\wedge\mathrm d^2x +\mathrm dQ\wedge\mathrm dy+Q\mathrm d^2y\\
		=&\left(P_x\mathrm dx+P_y\mathrm dy\right)\wedge\mathrm dx+\left(Q_x\mathrm dx+Q_y\mathrm dy\right)\wedge\mathrm dy\\
		=&P_x\mathrm d^2x+P_y\mathrm dy\wedge\mathrm dx+Q_x\mathrm dx\wedge\mathrm dy+Q_y\mathrm d^2y=(Q_x-P_y)\mathrm dx\wedge \mathrm dy
	\end{align*}
	代入$\int_{\partial D}\omega=\int_{D}\mathrm d\omega$得证.
\end{solution}

\begin{problem}
	已知光滑流形间的光滑映射$\varphi\colon M\to N$可以诱导其光滑微分$p$-形式间的拉回映射.
	\begin{itemize}
		\item 试问: 以下那个选项实拉回映射的正确表达形式.
		\begin{align*}
			(A)\quad\varphi^\star\colon \mathcal A^p(M)\to \mathcal A^p(N)&&(B)\quad\varphi^\star\colon \mathcal A^p(N)\to \mathcal A^p(M)
		\end{align*}
		\item 试证明: 外积运算$\wedge$与拉回映射$\varphi^\star$可交换.
	\end{itemize}
\end{problem}

\begin{solution}
	\begin{itemize}
		\item (B)
		\item 对任意$\alpha=\alpha^1\wedge\cdots\wedge\alpha^p\in\mathcal A^p(N)$, $\beta=\beta^1\wedge\cdots\wedge\beta^q\in\mathcal A^q(N)$.
		\begin{align*}
		\varphi^\star(\alpha\wedge\beta)=&\varphi^\star(\alpha^1\wedge\cdots\wedge\alpha^p\wedge\beta^1\wedge\cdots\wedge\beta^q)\\
		=&\varphi^\star(\alpha^1)\wedge\cdots\wedge\varphi^\star(\alpha^p)\wedge\varphi^\star(\beta^1)\wedge\cdots\wedge\varphi^\star(\beta^q)=\varphi^\star(\alpha)\wedge\varphi^\star(\beta)
		\end{align*}
	\end{itemize}
\end{solution}

\begin{problem}
	实系数$2$阶一般线性群定义为
	\begin{align*}
		\operatorname{GL}(2,\mathbb R)=\left\{A=\begin{bmatrix}
			a_1^1&a_1^2\\
			a_2^1&a_2^2
		\end{bmatrix}\colon \operatorname{det}A\neq 0,\quad a_i^j\in\mathbb R,\quad i,j=1,2\right\},
	\end{align*}
	其中$\operatorname{det}A$为$2\times 2$矩阵$A$的行列式.
	\begin{itemize}
		\item 试证明: $\operatorname{GL}(2,\mathbb R)$关于矩阵乘法为一个Lie群.
		\item 试求: $\operatorname{GL}(2,\mathbb R)$上, 在单位矩阵$I_2$处切空间$\operatorname T_{\operatorname{GL}(2,\mathbb R),I_2}$上取值为$\left.\dfrac{\partial}{\partial a_1^1}\right|_{I_2}$的左不变向量场$\omega_1^1$.
	\end{itemize}
\end{problem}
\begin{solution}
	\begin{itemize}
		\item $\operatorname{GL}(2,\mathbb R)$是一个$4$维光滑流形, 且其关于矩阵乘法是一个群. 其满足映射\begin{align*}
			\operatorname{GL}(2,\mathbb R)\times \operatorname{GL}(2,\mathbb R)\to \operatorname{GL}(2,\mathbb R),(A,B)\mapsto AB^{-1}.
		\end{align*}是光滑的, 从而证明$\operatorname{GL}(2,\mathbb R)$关于矩阵乘法为一个Lie群.
		其中只需证明$\operatorname{GL}(2,\mathbb R)$确实是一个$4$维光滑流形即可.
		
		用$\operatorname M_2$表示$2\times 2$的实矩阵全体, 对任意$A\in\operatorname M_2, A=\begin{bmatrix}
			a_1^1&a_1^2\\
			a_2^1&a_2^2
		\end{bmatrix}$, 令$\varphi(A)=(a_1^1, a_1^2,a_2^1,a_2^2)$. 其中$a_i^j\in\mathbb R$, 于是$\varphi$是$\operatorname M_2$到$\mathbb R^4$的一个$1$-$1$对应. 
		
		在$\operatorname M_2$上赋予拓扑, $\operatorname M_2$的一个集合$U$称为是开集当且仅当$\varphi(U)$是$\mathbb R^4$中的一个开集. 
		
		于是$\varphi$为$\operatorname M_2$到$\mathbb R^4$上的一个同胚. $(\operatorname M_2,\varphi)$就是覆盖$\operatorname M_2$的一个$C^\infty$-图册, 从而$\operatorname M_2$是一个$4$维微分流形.
		
		有$\operatorname{det}\colon M_2\to \mathbb R\in C^\infty$. 易知$\operatorname{GL}(2,\mathbb R)=\operatorname M_2\setminus\operatorname{det}^{-1}\{0\}$, 从而$\operatorname{GL}(2,\mathbb R)$是$\operatorname M_2$的一个开子集, 从而$(\operatorname{GL}(n,\mathbb R),\varphi)$是覆盖$\operatorname{GL}(2,\mathbb R)$的一个$C^\infty$-图册, 从而$\operatorname{GL}(2,\mathbb R)$是一个$4$维微分流形.
		\item $\operatorname{GL}(2,\mathbb R)$是$\operatorname M_2\simeq \mathbb R^4$的一个开子集, 所以$\operatorname{GL}(2,\mathbb R)$的(典范的)Lie代数的底空间, 即$\operatorname{GL}(2,\mathbb R)$在$I_2$处切空间$\operatorname T_{\operatorname{GL}(2,\mathbb R),I_2}$, 就是$M_2$自身, 即$\mathfrak{gl}(2,\mathbb R)=\operatorname M_2$.
		
		对任意$A=(A_i^j)\in\mathfrak{gl}(2,\mathbb R)$, 并记$\operatorname{GL}(2,\mathbb R)$上全局坐标系为$(a_i^j)$, 那么$A$相应的在$\operatorname T_{\operatorname{GL}(2,\mathbb R), I_2}$处的切向量就是$\sum A_i^j\dfrac{\partial}{\partial a_i^j}$. 这说明了$A=\begin{bmatrix}
			1&0\\0&0
		\end{bmatrix}$. 而其生成的左不变场在$X=(a_i^j)$处向量为$\sum a_i^kA_k^j\dfrac{\partial}{\partial a_i^j}=\sum\limits_ia_i^1\dfrac{\partial}{\partial a_i^1}$, 即$\omega_1^1=a_1^1\dfrac{\partial}{\partial a_1^1}+a_2^1\dfrac{\partial}{\partial a_2^1}$.
	\end{itemize}
\end{solution}


\end{document}
